**Question:**

Find the direction in which a straight line must be drawn through the point $(-1,2)$ so that its point of intersection with the line $x+y=4$ may be at a distance of 3 units from this point.

**Solution:**

Let *y* = *mx* + *c* be the line through point (–1, 2).

Accordingly, 2 = *m* (–1) + *c*.

$\Rightarrow 2=-m+c$

$\Rightarrow c=m+2$

$\therefore y=m x+m+2 \ldots$ (1)

The given line is

$x+y=4 \ldots(2)$

On solving equations (1) and (2), we obtain

$x=\frac{2-m}{m+1}$ and $y=\frac{5 m+2}{m+1}$

$\therefore\left(\frac{2-m}{m+1}, \frac{5 m+2}{m+1}\right)$ is the point of intersection of lines (1) and (2).

Since this point is at a distance of 3 units from point (– 1, 2), according to distance formula,

$\sqrt{\left(\frac{2-m}{m+1}+1\right)^{2}+\left(\frac{5 m+2}{m+1}-2\right)^{2}}=3$

$\Rightarrow\left(\frac{2-m+m+1}{m+1}\right)^{2}+\left(\frac{5 m+2-2 m-2}{m+1}\right)^{2}=3^{2}$

$\Rightarrow \frac{9}{(m+1)^{2}}+\frac{9 m^{2}}{(m+1)^{2}}=9$

$\Rightarrow \frac{1+m^{2}}{(m+1)^{2}}=1$

$\Rightarrow 1+m^{2}=m^{2}+1+2 m$

$\Rightarrow 2 m=0$

$\Rightarrow m=0$

Thus, the slope of the required line must be zero i.e., the line must be parallel to the *x*-axis.